What is the value of $c$ if $x\cdot(3x+1)<c$ if and only when $x\in \left(-\frac{7}{3},2\right)$?
Solution: When $x\in \left(-\frac{7}{3},2\right)$, we have $x\cdot(3x+1)-c<0$. This means that $x(3x+1)-c=0$ at $x=-\frac{7}{3}$ and $x=2$. We now know that $x(3x+1)-c=0$ is a quadratic equation with roots at $x=-\frac{7}{3}$ and $x=2$, and we want to use these roots to find a quadratic of the same form as the problem.  $x=-\frac{7}{3}$ gives us $(3x+7)=0$ and $x=2$ gives us $(x-2)=0$. \begin{align*}
x(3x+1)-c&=(3x+7)(x-2)\\
&=3x^2+x-14\\
&=x(3x+1)-14.
\end{align*} So, $c=\boxed{14}$.